Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks for the number of positive integers n, such that the factorial of n ends with exactly m zeroes. Are you among those great programmers who can solve this problem?
Input The only line of input contains an integer m (1 ≤ m ≤ 100 000) — the required number of trailing zeroes in factorial.
Output First print k — the number of values of n such that the factorial of n ends with m zeroes. Then print these k integers in increasing order.
Example Input 1 Output 5 5 6 7 8 9 Input 5 Output 0 Note The factorial of n is equal to the product of all integers from 1 to n inclusive, that is n! = 1·2·3·…·n.
In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880. 题意：让你找N的阶乘的末尾有m个0的数有多少，并输出n. 解法：因为末尾是0和5的个数有关系，n！=n*（n-1）（n-2）…….*1，里面有多少个5，那么结果的末尾就有多少个0
#include<stdio.h> #include<string.h> int a[10000]; int main() { int n; while(scanf("%d",&n)!=-1) { int i,j; for(i=1;n>0;i++)//1*2*3*.....*(n-1)*找能被5整除的数 { int we=i; while(we%5==0)//找到了就零的个数-- { we /= 5; n--; } } if(n<0) { printf("0\n"); } else { printf("5\n"); for(int j=i-1;j<=i+3;j++) { if(j<i+3) printf("%d ",j); if(j==i+3) printf("%d\n",j); } } } }